Please explain these counter-interpretations to these Natural Deduction arguments

Question

I have attached a screenshot of the arguments, but here they are written out just in case. I am able to prove when sentences are equivalent, but could somebody please explain the counter-interpretations to me?

I understand that, since its a counterinterpretation, they have opposite truth-values, but I don't understand how the use of numbers reflect truth-values. For example, in the first question, since "" names 1, how do I understand "" ? How does the predicate act upon "", and what is the significance of "" symbolising 1? Also, still in the first question, there is no in either sentence, so what relevance does it have to the question? I am unsure of the use of numbers as counterexamples in general. I hope that clarifies my question.

Argument: $∀xPx → Qc,∀x(Px → Qc)$ Counter-interpretation: let the domain be the numbers 1 and 2. Let $‘c’$ name 1. Let $‘Px’$ be true of and only of 1. Let $‘Qx’$ be true of, and only of, 2.

Argument: $∀x∀y∀zBxyz$,$∀xBxxx$ Counter-interpretation: let the domain be the numbers 1 and 2. Let ‘$Bxyz$’ be true of, and only of, ⟨1,1,1⟩ and ⟨2,2,2⟩.

Argument: $∃x∀yDxy$,$∀y∃xDxy$ Counter-interpretation: let the domain be the numbers 1 and 2. Let $‘Dxy’$ hold of and only of ⟨1,2⟩ and ⟨2,1⟩. This is depicted thus: 1 ↔ 2

Argument: $∀x(Rca ↔ Rxa)$, $Rca ↔ ∀xRxa$ Counter-interpretation: consider the following diagram, allowing $‘a’$ to name 1 and ‘c’ to name 2: The diagram is of 1 with an arrow from itself pointing to itself, and 2 with no arrows.

Answer 1

For example, in the first question, since "c" names 1, how do I I don't understand how the use of numbers reflect truth-values. For example, in the first question, since "c" names 1, how do I understand "Qc" ? How does the predicate Q act upon "c", and what is the significance of "c" symbolising 1? Also, still in the first question, there is no Qx in either sentence, so what relevance does it have to the question? I am unsure of the use of numbers as counterexamples in general. I hope that clarifies my question

I'll elaborate on the first two given solutions by way of answering the above.

Argument: $$∀xPx → Qc,\quad∀x(Px → Qc)$$

Note that the screenshot doesn't actually use word "argument", and your introducing it during transcribing is a mistake: the given is merely a pair of sentences to be compared, rather than to be joined by a conditional (namely, an argument).

Counter-interpretation: let the domain be the numbers $1$ and $2.$

• Each variable (e.g., $x$) is an element of $\{1,2\}.$

Let ‘$c$’ name $1.$

• Weird phrasing; I presume that $c$ is a constant of value $1.$

Let ‘$Px$’ be true of and only of $1.$ Let ‘$Qx$’ be true of, and only of, $2.$

• $P=\{1\}, Q=\{2\};$ that is, $P1,Q2$ are true and $P2,Q1$ false. $P2$ is commonly written as $P(2)$.

Hence, $$∀xPx → Qc$$ becomes False→False, while putting $x{=}1$ into $Px → Qc$ shows that $$∀x(Px → Qc)$$ is true/false?

Argument: $$∀x∀y∀zBxyz,\quad∀xBxxx$$ Counter-interpretation: let the domain be the numbers $1$ and $2.$ Let ‘$Bxyz$’ be true of, and only of, $⟨1,1,1⟩$ and $⟨2,2,2⟩.$

• $x,y,z\in\{1,2\},\quad B=\{(1,1,1),(2,2,2)\}.$

Left-hand side: put $(x,y,z){=}(1,2,2).$

Right-hand side: can $Bxxx$ be falsified?